5.1 Calculus - SL content

A limit describes the value that $f(x)$ approaches as $x$ approaches a point. At SL, estimate it from a graph or from table values on both sides; formal analytic limit calculations are not required.

The derivative is the gradient function and an instantaneous rate of change. Notation identifies the changing quantities: $f'(x)$ or $dy/dx$ for a function, $dV/dr$ for volume changing with radius, and $ds/dt$ for displacement changing with time.

Example

If table values of $f(x)$ are 3.98 at $x=1.99$ and 4.02 at $x=2.01$, they support $\lim_{x\to2}f(x)\approx4$. If the tangent gradient there is 5 metres per second, the derivative gives the local rate and its units.

The function value and limit need not agree when there is a hole, and one-sided behaviour may disagree at a jump. Do not substitute $x=2$ blindly; inspect values approaching from both sides.

Where f'(x)>0, f is increasing locally; where f'(x)<0, f is decreasing locally. A stationary point has f'(x)=0, but the sign must be checked on either side to classify the change.

Find critical x-values, split the domain into intervals and test the derivative sign. This produces a variation table and keeps domain restrictions visible instead of relying on a sketch.

For f(x)=x³−3x, f'(x)=3x²−3. The derivative is positive outside x=−1 and x=1 and negative between them, so the function rises, falls, then rises again.

f'(x)=0 does not automatically mean a maximum or minimum. A horizontal inflection can have zero derivative without changing from increasing to decreasing.

For $f(x)=ax^n$ with integer $n$, $f'(x)=anx^{n-1}$. Differentiate a sum term by term and treat a constant as having derivative zero.

Negative integer powers are included where the original function is defined: $d(x^{-2})/dx=-2x^{-3}$. Keep coefficients, signs and domain restrictions visible before simplifying.

Example

For $f(x)=4x^3-2x^{-1}+7$, $f'(x)=12x^2+2x^{-2}=12x^2+2/x^2$, with $x\ne0$ inherited from the original function.

This SL rule is bounded to integer exponents. Rational powers, chain, product and quotient rules belong to AHL 5.9; do not import them into this Objective.

At x=a, the tangent gradient is f'(a). If that gradient is non-zero, the normal gradient is −1/f'(a), because perpendicular non-vertical lines have product of gradients −1.

Find the point (a,f(a)), calculate the tangent gradient, then use point–gradient form. Handle a horizontal tangent separately: its normal is vertical and cannot be written with a finite gradient.

For y=x² at x=1, the point is (1,1), tangent gradient 2 and tangent y−1=2(x−1). The normal gradient is −1/2, giving y−1=−(x−1)/2.

A normal is not the negative of the tangent gradient. It is the negative reciprocal, and the vertical/horizontal special cases must be stated.

For integer $n\ne-1$, $\int ax^n\,dx=ax^{n+1}/(n+1)+C$. An indefinite integral is a family of anti-derivatives; a boundary condition determines the constant $C$.

A definite integral $\int_a^b f(x)\,dx$ connects anti-derivatives with accumulated signed area. At SL, use technology after first writing the correct integral; for a region above the x-axis, the integral equals its geometric area.

Example

If $dy/dx=3x^2+x$ and $y=10$ when $x=1$, then $y=x^3+x^2/2+C$ and $10=1+1/2+C$, so $C=8.5$. Thus $y=x^3+x^2/2+8.5$.

Do not omit $C$ in an indefinite integral or apply the power formula to $x^{-1}$. A differential-equation growth model belongs to AHL, not this SL anti-differentiation Objective.

A stationary point occurs at an interior value where $f'(x)=0$. Use technology when appropriate to generate $f'(x)$ and solve for the x-values, then calculate the corresponding y-values.

Classify a local maximum when the function changes from increasing to decreasing and a local minimum when it changes from decreasing to increasing. Compare endpoints as well when the greatest or least value on a restricted domain is required.

Example

For $f(x)=x^3-3x$, $f'(x)=3x^2-3=0$ at $x=\pm1$. The derivative changes $+$ to $-$ at $-1$ and $-$ to $+$ at $1$, giving a local maximum and minimum respectively.

$f'(x)=0$ alone does not guarantee an extremum, and a local extremum need not be the absolute extremum on the stated domain. The second-derivative test belongs to AHL 5.10.

Optimization maximises or minimises a specified objective subject to constraints. Calculus can locate interior candidates where the derivative is zero, but endpoints, feasibility and the meaning of the variables decide the answer.

Translate the context into an objective function, state the domain, solve for critical points and compare all feasible candidates. A minimum cost or maximum area is a claim about the whole allowed interval, not just a local curve shape.

For a rectangle with fixed perimeter 20, A=x(10−x) on 0<x<10. A'(x)=10−2x gives x=5, and comparing the endpoints' limiting values confirms the square gives the largest area.

A stationary point is not automatically the global optimum. Check the full feasible domain and endpoints, state units, and interpret the best value in context. Kinematics questions are not set in SL examinations.

For equally spaced x-values with width h, the trapezoidal rule approximates ∫ₐᵇ f(x)dx by h/2[f(x₀)+2f(x₁)+⋯+2f(xₙ₋₁)+f(xₙ)]. It estimates signed area, not automatically total geometric area.

The endpoint ordinates have weight one and interior ordinates weight two. More strips usually reduce error for a smooth curve, but curvature and the number of strips still determine the approximation quality.

Using two strips for f(x)=x² on [0,2] gives h=1 and (1/2)[0+2(1)+4]=3, while the exact integral is 8/3. The difference is the discretisation error.

Do not use the rule with the wrong h or omit the doubled interior values. If the curve crosses the axis, signed integral and total area require different treatment.